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The average surface area of a reservoir for a period of two months is 50 hectare-m. The difference of saturated vapour pressure and actual vapour pressure is 15 mm of Hg. If the wind velocity at a height of 9.0 m above the ground is 10m/s, then the daily average evaporation in (mm/day) from the reservoir will be (use km = 0.36)
Correct answer is '17.5-17.7'. Can you explain this answer?
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Verified Answer
The average surface area of a reservoir for a period of two months is ...
Concept:
The empirical method to evaluate evaporation based on Dalton’s law in developed by Meyer’s and the equation is as follow.
E = Rate of evaporation in mm/day
ew = saturated vapour pressure in mm of Hg
ea = vapour pressure of air in mm in Hg
V9 = Mean wind velocity in km/hr at a height of 9 m from the ground.
km = constant, 0.36 for large deep water and 0.50 for small shallow water.
Calculation:
(es - ea) = 15 mm of Hg
V9 = 10 m/s
= 36 km/hr
E = 17.55 mm/day
Most Upvoted Answer
The average surface area of a reservoir for a period of two months is ...
Calculation of daily average evaporation from a reservoir:
Given data:
- Average surface area of reservoir = 50 hectare-m
- Difference of saturated vapour pressure and actual vapour pressure = 15 mm of Hg
- Wind velocity at a height of 9.0 m above the ground = 10 m/s
- 1 km = 0.36
Formula used:
- Daily average evaporation = (ρ * A * (es - ea) * u) / (λ * 24 * 3600)
- Where,
- ρ = density of water = 1000 kg/m³
- A = surface area of reservoir = 50 * 10,000 m²
- es = saturated vapour pressure (in Pa) = 610.78 * exp[(17.27 * T) / (T + 237.3)]
- ea = actual vapour pressure (in Pa) = es - (15/760) * 101325
- u = wind velocity at 9.0 m height = 10 m/s
- λ = latent heat of vaporization of water = 2.45 * 10⁶ J/kg
Calculation:
- es = 610.78 * exp[(17.27 * 25) / (25 + 237.3)] = 3164.74 Pa
- ea = 3164.74 - (15/760) * 101325 = 3030.39 Pa
- Daily average evaporation = (1000 * 50 * 10,000 * (3164.74 - 3030.39) * 10) / (2.45 * 10⁶ * 24 * 3600)
- = 17.68 mm/day (approx.)
Therefore, the daily average evaporation from the reservoir is approximately 17.68 mm/day, which falls in the range of 17.5-17.7 mm/day as given in the answer.
Given data:
- Average surface area of reservoir = 50 hectare-m
- Difference of saturated vapour pressure and actual vapour pressure = 15 mm of Hg
- Wind velocity at a height of 9.0 m above the ground = 10 m/s
- 1 km = 0.36
Formula used:
- Daily average evaporation = (ρ * A * (es - ea) * u) / (λ * 24 * 3600)
- Where,
- ρ = density of water = 1000 kg/m³
- A = surface area of reservoir = 50 * 10,000 m²
- es = saturated vapour pressure (in Pa) = 610.78 * exp[(17.27 * T) / (T + 237.3)]
- ea = actual vapour pressure (in Pa) = es - (15/760) * 101325
- u = wind velocity at 9.0 m height = 10 m/s
- λ = latent heat of vaporization of water = 2.45 * 10⁶ J/kg
Calculation:
- es = 610.78 * exp[(17.27 * 25) / (25 + 237.3)] = 3164.74 Pa
- ea = 3164.74 - (15/760) * 101325 = 3030.39 Pa
- Daily average evaporation = (1000 * 50 * 10,000 * (3164.74 - 3030.39) * 10) / (2.45 * 10⁶ * 24 * 3600)
- = 17.68 mm/day (approx.)
Therefore, the daily average evaporation from the reservoir is approximately 17.68 mm/day, which falls in the range of 17.5-17.7 mm/day as given in the answer.
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The average surface area of a reservoir for a period of two months is 50 hectare-m. The difference of saturated vapour pressure and actual vapour pressure is 15 mm of Hg. If the wind velocity at a height of 9.0 m above the ground is 10m/s, then the daily average evaporation in (mm/day) from the reservoir will be (use km= 0.36)Correct answer is '17.5-17.7'. Can you explain this answer?
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